Question

An iceberg of volume 2100cm³ and density0.8g/cm³ is floating in the sea water of density 1.2g/cm³. Then the volume of iceberg immersed in water is​

Answer 1

Answer:

your answer

Step-by-step explanation:

lets say V is the total volume of the iceberg, and x be the volume of submerged part of iceberg. then according to Archimedes principle, we have

weight of the iceberg= weight of the water displaced (weight of x volume of water)

⇒V×0.92×g=x×1.03×g

⇒

V

x

=

103

92

percentage of volume of iceberg submerged in sea water =

103

92

×100=89.32%

so percentage of total volume of iceberg above the level of sea water =100−89.32=10.68%≈11%

Answer 2

Answer:

The volume of iceberg immersed in water is​ 1400cm³.

Step-by-step explanation:

Given the volume of iceberg, $V_i = 2100cm^3$

The density of iceberg, $d_i = 0.8g/cm^3$

The density of water, $d_w = 1.2g/cm^3$

According to Archimedes principle, the volume of the submerged object is equal to the volume of the fluid displaced.

Since density is given mass occupied per unit volume,

$d=\dfrac{m}{V} \implies m=dV$

therefore we can say that mass of the iceberg is equal to the mass of the water displaced. i.e.,

$V_i\times d_i=V_w\times d_w$

$\implies 2100 \times 0.8=V_w\times 1.2$

$\implies V_w =\dfrac{2100 \times 0.8}{1.2} =1400cm^3$

Volume of iceberg immersed in water is nothing but the volume of water displaced.

Therefore, the volume of iceberg immersed in water is​ 1400cm³.

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